Simplify the following expression and state the condition under which the simplification is valid. $q = \dfrac{-9a^3 - 18a^2 + 135a}{a^3 + 10a^2 + 25a}$
Solution: First factor out the greatest common factors in the numerator and in the denominator. $ q = \dfrac {-9a(a^2 + 2a - 15)} {a(a^2 + 10a + 25)} $ $ q = -\dfrac{9a}{a} \cdot \dfrac{a^2 + 2a - 15}{a^2 + 10a + 25} $ Simplify: $ q = - 9 \cdot \dfrac{a^2 + 2a - 15}{a^2 + 10a + 25}$ Since we are dividing by $a$ , we must remember that $a \neq 0$ Next factor the numerator and denominator. $ q = - 9 \cdot \dfrac{(a + 5)(a - 3)}{(a + 5)(a + 5)}$ Assuming $a \neq -5$ , we can cancel the $a + 5$ $ q = - 9 \cdot \dfrac{a - 3}{a + 5}$ Therefore: $ q = \dfrac{ -9(a - 3)}{ a + 5 }$, $a \neq -5$, $a \neq 0$